Question

Indicate the oxidation number of phosphorous in the following ions. 

a) PO₄^3- b) PO₃^3- c) HPO₄^2- d) P₃O₁₀^5-

Answer 1

Ans-- For calculating oxidation number of en element in a coumpounds we have to follow some rules as follows:

a) given compound is PO₄^3- , we have to calulate the oxidation number of p using above rules .

let oxidation number of phosphorus is x , then we can write

X+4(-2)=-3

X-8=-3

X=-3+8

X=+5

Hence , phosphorus has a +5 oxidation state in the phosphate anion, PO3−4

1. The oxidation number of a free element is always 0.

2. The oxidation number of a monatomic ion equals the charge of the ion.

3. The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.

4. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.

5. The oxidation number of a Group 1 element in a compound is +1.

6. The oxidation number of a Group 2 element in a compound is +2.

7. The oxidation number of a Group 17 element in a binary compound is -1.

8. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

b) PO₃^3-

^{We can write ,} X+3(-2)=-3

X-6=-3

X=+3

Hence , phosphorus has a +3 oxidation state

C) HPO₄^2-

We can write as , 1+ x+ 4(-2)=-2

1+x-8=-2

x=-2+8-1

x=+5

Hence , phosphorus has a +5 oxidation state

d) given compound is P_3^O10^5-

3(X)+10(-2)=-5

3X-20=-5

3X=-5+20

3X=15

X=15/3

X=+5

Hence , phosphorus has a +5 oxidation state